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SOVA2 [1]
3 years ago
9

A fairgrounds ride spins its occupants inside a flying saucer-shaped container. If the horizontal circular path the riders follo

w has a 9.00 m radius, at how many revolutions per minute will the riders be subjected to a centripetal acceleration whose magnitude is 1.80 times that due to gravity
Physics
1 answer:
cestrela7 [59]3 years ago
4 0

Answer:

13.37 rev/min

Explanation:

acceleration due to gravity (g) = 9.8 m/s², centripetal acceleration (a_c) = 1.8 * g = 1.8 * 9.8 m/s² = 17.64 m/s².

r = 9 m

Centripetal acceleration (a_c) is given by:

a_c=\frac{v^2}{r} \\\\v=\sqrt{a_c*r} \\\\v=\sqrt{17.64\ m/s^2*9\ m}\\\\v=12.6\ m/s

The velocity (v) is given by:

v = ωr;  where ω is the angular velocity

Hence:

ω = v/r = 12.6 / 9

ω = 1.4 rad/s

ω = 2πN

N = ω/2π = 1.4 / 2π

N = 0.2228 rev/s

N = 13.37 rev/min

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Answer: A student walks 50 meters east, 40 meters north, 35 meters east, and then 20 m south. Then the magnitude and direction of the student's total displacement will be 87.32 m along the direction of AD or in east-south direction.

Explanation: To find the correct answer, we need to know about the Displacement of a body in motion.

<h3>What is displacement of a body in motion?</h3>
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AD=AC+CD\\AC^{2} =50^{2} +11^2\\AC=51.19 m\\Similarly,\\CD^2=35^2+9^2\\CD=36.13 m\\thus, \\AD=51.19+36.13=87.32 m

Thus, from the above calculations, we can conclude that, the displacement of the body will be equal to 87.32 m along the direction of AD or in east-south direction.

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